A Lie algebra over $\mathbb Z$ is defined to be an abelian group with a bracketing operation $[\cdot,\cdot]$, satisfying the Jacobi identity and the relation $[x,x]=0$ for every $x$. On the other hand, a quasi-Lie algebra is defined by replacing the axiom $[x,x]=0$ with the antisymmetry axiom $[x,y]+[y,x]=0$ for all $x$ and $y$. Jerry Levine used the free quasi-Lie algebra to study the group of homology cylinders. See this paper and this paper. My question is whether there are other contexts in which quasi-Lie algebras appear. Lie algebras arise naturally in all sorts of different contexts,
but I don't know of many places in which quasi-Lie algebras appear. For example, if one looks at the associated graded object for the lower central series of a group, one naturally obtains a Lie algebra and not a quasi-Lie algebra.

One possible answer is this MO post, where it is noted that in operad language, the quasi-Lie axioms are necessary, and this is quite germane to why Levine needed to use quasi-Lie algebras, since he was essentially building a group out of the Lie operad. But I'd like to know of any other examples where quasi-Lie algebras occur that people might know.

Added (Moskovich): I wonder also whether Levine was the first person ever to consider quasi-Lie algebras, in 2002. Surely that can't possibly be the case- as a structure, it's surely too natural to be so new!

Added (Conant) This is not just an idle question. My main motivation for asking this question is to find new link concordance invariants. Milnor's mu invariants are associated to the lower central series of a link group, the associated graded object being a Lie algebra. I'm hoping there is some kind of quasi-Lie algebra one can assign to a link complement that generalizes this associated graded Lie algebra.

Edit: Based on the comments and Mariano's answer, I'd like to make some clarifications. First, when working over the integers, maybe it is better to use the term "Lie ring" as opposed to "Lie algebra," but I had just been following Levine's convention. (Unfortunately, Hilton's definition of a quasi-Lie ring, given in the reference in Mariano's answer, does not match Levine's!) Levine's definition means "a graded (or super) Lie algebra where all elements are of even degree." So one could take the even degree part of the Lie algebra formed from homotopy groups under the Whitehead product, as Tom mentions.

Edit: It turns out that Levine's definition of a quasi-Lie algebra is a specific case of Hilton's definition (see Mariano's answer), where the module $M$ is $0$.

Just as a matter of terminology, “…one naturally obtains a Lie algebra and not a quasi-Lie algebra” seems unfair, since any Lie algebra is also quasi-Lie, surely? I this is sort of opposite to the usage of “non-abelian” to mean “not-necessarily-Abelian”; you're using “quasi-Lie” sometimes to mean “quasi-Lie but not Lie”? Quibbles aside, I do like this question: interesting concepts should always be demonstrated with natural examples!
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Peter LeFanu LumsdaineOct 19 '10 at 16:48

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@Torsten: actually, I changed my mind. That's not cheating. You can certainly get nice quasi-Lie algebras by tensoring graded Lie algebras with $\mathbb Z/2\mathbb Z$. But it would be nice to have examples over $\mathbb Z$ which contain something besides $2$-torsion.
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Jim ConantOct 19 '10 at 23:07

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As @Torsten Ekedahl said, the what you call a "quasi-Lie algebra" translates much better than what you call a "Lie algebra" to contexts like Super Vector Spaces. Super groups, and hence "super Lie algebras" definitely do arise in nature, and super Lie algebras are the super version of "quasi-Lie algebras", not of "Lie algebras". Another example in super land is that the (super) commutator in an associative algebra satisfies the (super) "quasi-Lie" axiom but not the (super) "Lie" axiom.
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Theo Johnson-FreydOct 20 '10 at 1:33

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The homotopy groups of a (say, simply connected) space $X$ form a graded Lie algebra under Whitehead product, in which the even-dimensional part (which is actually the $\pi_n(X)$ for odd $n$) can have elements $x$ such that $[x,x]$ has order $2$ -- for example $\pi_n(S^n)$ for most odd values of $n$.
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Tom GoodwillieMar 10 '11 at 2:47

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@Tom: you probably didn't put this in an answer since it is awfully simple to be getting reputation points for, but it is a genuine answer...
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Jim ConantMar 10 '11 at 11:53

Unfortuntately, the term "quasi-Lie ring" used in this reference does not mean the same thing as Levine's term. However, the odd homotopy groups with Whitehead bracket does form a quasi-Lie algebra under Levine's definition.
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Jim ConantMar 10 '11 at 11:51

The homotopy groups of a (say, simply connected) space $X$ form a graded Lie algebra under Whitehead product, in which the even-dimensional part (which is actually the $\pi_n$ for odd $n$ ) can have elements $x$ such that $[x,x]$ has order $2$ -- for example $\pi_n(S^n)$ for most odd values of $n$.

Tom gave a perfectly fine answer, so I'm posting it as a community wiki and accepting it.
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Jim ConantMar 13 '11 at 1:54

@Jim: maybe there is something I am not seeing... but Hilton's motivating example is precisely the homotopy groups with the Whitehead bracket. Since he wants to consider all the groups (and since he is writing before super became a word), he defines quasi-Lie rings as pairs $(A,M)$, which in the case of the homotopy groups end up being $A$ the odd homotopy groups and $M$ the even homotopy groups, but it is clear that if $(A,M)$ is such an object, then $(A,0)$ is too, and that is the same thing as considering only the odd groups.
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Mariano Suárez-Alvarez♦Mar 13 '11 at 7:19

@Mariano: Thanks for the dictionary translating Hilton's terms into modern ones. I have to admit I didn't read it carefully, mainly because he finds both 2 and 3-torsion in the free quasi Lie ring, whereas there is only 2-torsion in Levine's free quasi-Lie algebra. So I concluded they are not the same definition. I think you are right that taking $M=0$ in Hilton's definition does give you a quasi Lie algebra in Levine's sense, although I'm still unsure where that 3-torsion is coming from.
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Jim ConantMar 13 '11 at 13:33

@Jim: I suspect the 3-torsion is coming from 3[x,[x,x]]=0 by the Jacobi relation.
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Daniel MoskovichMar 31 '11 at 20:43

@Daniel: But $[x,[x,x]]=0$ by the Jacobi identity! I guess in the super case, if $x$ has odd degree, this would be $3$-torsion.
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Jim ConantMar 31 '11 at 21:30